If both the speed and weight of the vehicle double, how much braking force is required to stop over the same distance?

To determine how much braking force is needed when both the speed and weight of a vehicle double, we can utilize the principles of physics pertaining to momentum and stopping distance.

When speed increases, the kinetic energy of the vehicle increases significantly, since kinetic energy is proportional to the square of the speed. If the initial speed is v, the kinetic energy is given by the formula ( KE = \frac{1}{2}mv^2 ). When the speed doubles (2v), the kinetic energy becomes ( KE = \frac{1}{2}m(2v)^2 = 2mv^2 ), which means it becomes four times the original kinetic energy (4 times).

Moreover, if the weight of the vehicle also doubles, this effectively doubles the mass (m). Therefore, the new kinetic energy can now be described as ( KE = \frac{1}{2}(2m)(2v)^2 = 4mv^2 ), which doubles the initial kinetic energy contribution from weight and quadruples the contribution from the speed.

To stop the vehicle over the same distance, the braking force must counteract this new total kinetic energy, which has increased to a factor of 8 compared to the original. This requirement makes it